The ground state energy of hydrogen atom is 13.6 eV. If an electron makes a transition from an energy level -1.51 {\eV} to -3.4 {\eV}, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.

CBSE Class 12 Physics 2017 Outside Delhi Set 3 Paper

Question : 4 of 8

Marks: +1, -0

The ground state energy of hydrogen atom is 13.6 eV. If an electron makes a transition from an energy level

-1.51 {\eV}

-3.4 {\eV}

, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.

Solution:

(a) Calculation of energy difference
(b) Formula
(c) Calculation of wavelength
(d) Name of the series of spectral lines
Energy difference

=3.4 {\eV}-1.51 {\eV}=1.89

{\eV}=3.024 × 10^{-19} {\J}

Energy

=\;{h c}/{λ}=3.024 × 10^{-19} {\J}

Wavelength

=6.57 × 10^{-7} {\m}

Series is Balmer series.

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